The dateline for this assignment is Wednesday

Assignment

1. (7.47) In his management information systems textbook, Professor David Kroenke raises an interesting point: “If 98% of our market has Internet access, do we have a responsibility to provide non-Internet materials to that other 2%? Suppose that 98% of the customers in your market do have Internet access, and you select a random sample of 500 customers. What is the probability that the sample has

a. Greater than 99% of the customers with internet access?

b. Between 97% and 99% of the customers with Internet access?

c. Fewer than 97% of the customers with Internet access?

2. (8.25) One operation of a mill is to cut pieces of steel into parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw, and the resulting parts must be cut to be within+/- 0.005 inch of the length specified by the automobile company. The measurement reported from a sample of 100 steel parts (stored in

Steel) is the difference, in inches, between the actual length of the steel part, as measured by a laser measurement device, and the specified length of the steel part. For example, the first observation, -0.002 represents a steel part that is 0.002 inch shorter than the specified length.

a. Construct a 95% confidence interval estimate for the population mean difference between the actual length of the steel part and the specified length of the steel part.

b. What assumption must you make about the population distribution in order to construct the confidence interval estimate in (a)?

c. Do you think that the assumption needed in order to construct the confidence interval estimate in (a) is valid? Explain.

3. (9.31) One operation of a steel mill is to cut pieces of steel into parts that are used in the frame for front seats in an automobile. The steel is cut with a diamond saw and requires the resulting parts must be cut to be within +/-0.005 inch of the length specified by the automobile company. The file

Steel

contains a sample of 100 steel parts. The measurement reported is the difference, in inches, between the actual length of the steel part, as measured by a laser measurement device, and the specified length of the steel part. For example, a value of -0.002 represents a steel part that is 0.002 inch shorter than the specified length.

a. At the 0.05 level of significance, is there evidence that the mean difference is not equal to 0.0 inches?

b. Construct a 95% confidence interval estimate of the population mean. Interpret this interval

c. Compare the conclusions reached in (a) and (b).

d. Because n = 100, do you have to be concerned about the normality assumption needed for the t test and t interval?

4. (9.33) Although many people think they can put a meal on the table in a short period of time, an article reported that they end of spending about 40 minutes doing so . Suppose another study is conducted to test the validity of this statement. A sample of 25 people is selected, and the length of time to prepare and cook dinner (in minutes) is recorded with the following results in (

Dinner

): sample of 50 cans, and the mean amount of paint per 1-gallon can is 0.995 gallon.

44.0 51.9 49.7 40.0 55.5 33.0 43.4 41.3 45.2 40.7 41.1 49.1 30.9

45.2 55.3 52.1 55.1 38.8 43.1 39.2 58.6 49.8 43.2 47.9 46.6

a. Is there evidence that the population time to prepare and cook dinner is different from 40 minutes? Use the p-value approach and a level of significance of 0.05.

b. What assumption about the population distribution is needed in order to conduct the t test in (a)?

c. Make a list of the various ways you could evaluate the assumption noted in (b).

d. Evaluate the assumption in (b) and determine whether the t test in (a) is valid.

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